First, find the concentration of the NaCN solution in terms
of molarity. The molar mass of NaCN is 49.0072 g/mol
Molarity = mol solute ÷ L solution
Molarity = (353 mg * (1 g / 1000 mg) * (1 / 49.0072 g/mol)) ÷ (500 mL * (1 L / 1000 mL))
Molarity = 0.0144 M NaCN
NaCN is a salt from a weak acid HCN and a strong base NaOH.
The reaction is as follows:
HCN + NaOH --> NaCN + H2O
Upon hydrolysis, it becomes:
+ --> HCN +
Initial: Amount of is 0.0144 M, is in excess, and
the products are 0.
Change: This is unknown. Let this be x. For the , it
would be –x, while it would be +x for the products HCN and .
Excess: This would be Initial + Change. Thus for , it
would be 0.0144 – x. For both products, it would be x for each.
The hydrolysis constant () would be the amount of products
over the reactants. Thus,
= (HCN)*() / () = x2 / (0.0144 –x) --> equation 1
is also equal to the equilibrium constant of water ( =
1 x 10^-14) divided by .
= 1 x ÷ 2.1 x --> equation 2
Substituting equation 2 to equation 1:
÷ (0.0144 –x) = 4.762 x
Solving for x,
X = 2.59 x
As mentioned before, x denotes the products HCN and .
Thus the amount of is 2.59 x
To find pH,
pH = 14-(-log )
pH = 14 – (-log 2.59 x )
pH = 10.41
Thus, the answer is 10.41.